A community for students.
Here's the question you clicked on:
 0 viewing
Sujay
 3 years ago
You are a farmer, and you are trying to make the most efficient use of your resources. You have invested enough capital for a barn house which is 40 feet by 70 feet (its base) and 100 feet of fencing for your animals. You decide to have an outdoor lot for your animals. Your goal is to have the lot with the most area with the least amount of fencing (optimize). The lot must be rectangular, and one side of the rectangular may be a wall from the barn. A picture will be posted below.
Sujay
 3 years ago
You are a farmer, and you are trying to make the most efficient use of your resources. You have invested enough capital for a barn house which is 40 feet by 70 feet (its base) and 100 feet of fencing for your animals. You decide to have an outdoor lot for your animals. Your goal is to have the lot with the most area with the least amount of fencing (optimize). The lot must be rectangular, and one side of the rectangular may be a wall from the barn. A picture will be posted below.

This Question is Closed

AnimalAin
 3 years ago
Best ResponseYou've already chosen the best response.1The area of your pen is the length times the width. Since you are going to have three sides of the pen enclosed by your fencing (the last side will be from the barn), we have a second specification that must be met, the length plus twice the width is 100. Rearrange the second equation to solve for a single variable (either l or w) and substitute into the first equation. You will get a quadratic function; find the maximum value, and you have the width (or maybe length) that corresponds to the maximum area. One more substitution gives the dimensions of the pen..

AnimalAin
 3 years ago
Best ResponseYou've already chosen the best response.1\[A=lw\]\[100=l+2w \implies l=1002w \implies\]\[A=lw=A(w)=(1002w)w=2w^2+100\]To find the maximum (or minimum in other cases) of a quadratic function f(x)=ax^2+bx+c, we take the value of the function at x=b/(2a). In this case, that gives us\[w=\frac{100}{2(2)}=25 \implies l=50\]by substitution. So our enclosure is 25' by 50', against the side of the barn that is 70' long.

Sujay
 3 years ago
Best ResponseYou've already chosen the best response.0Wow thanks, I totally forgot we don't need calculus for this problem.

Sujay
 3 years ago
Best ResponseYou've already chosen the best response.0Can anyone do it using optimization?

AnimalAin
 3 years ago
Best ResponseYou've already chosen the best response.1The optimization problem is the same, except when you get A(w), you take the derivative and set it to zero. Since the second derivative is negative at that value of w, the function has a maximum at that point.\[A(w)=−2w^2+100w \implies A'(w)=4w+100=0 \implies w=25\]\[A"(w)=4<0 \implies maximum~at~critical~value\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.